Django: add image in an ImageField from image url

please excuse me for my ugly english ;-)

Imagine this very simple model :

class Photo(models.Model):
    image = models.ImageField('Label', upload_to='path/')

I would like to create a Photo from an image URL (i.e., not by hand in the django admin site).

I think that I need to do something like this :

from myapp.models import Photo
import urllib

img_url = 'http://www.site.com/image.jpg'
img = urllib.urlopen(img_url)
# Here I need to retrieve the image (as the same way that if I put it in an input from admin site)
photo = Photo.objects.create(image=image)

I hope that I've well explained the problem, if not tell me.

Thank you :)

Edit :

This may work but I don't know how to convert content to a django File :

from urlparse import urlparse
import urllib2
from django.core.files import File

photo = Photo()
img_url = 'http://i.ytimg.com/vi/GPpN5YUNDeI/default.jpg'
name = urlparse(img_url).path.split('/')[-1]
content = urllib2.urlopen(img_url).read()

# problem: content must be an instance of File
photo.image.save(name, content, save=True)

I just created http://www.djangosnippets.org/snippets/1890/ for this same problem. The code is similar to pithyless' answer above except it uses urllib2.urlopen because urllib.urlretrieve doesn't perform any error handling by default so it's easy to get the contents of a 404/500 page instead of what you needed. You can create callback function & custom URLOpener subclass but I found it easier just to create my own temp file like this:

from django.core.files import File
from django.core.files.temp import NamedTemporaryFile

img_temp = NamedTemporaryFile(delete=True)
img_temp.write(urllib2.urlopen(url).read())
img_temp.flush()

im.file.save(img_filename, File(img_temp))

from myapp.models import Photo
import urllib
from urlparse import urlparse
from django.core.files import File




img_url = 'http://www.site.com/image.jpg';




photo = Photo()    # set any other fields, but don't commit to DB (ie. don't save())
name = urlparse(img_url).path.split('/')[-1]
content = urllib.urlretrieve(img_url)



See also: http://docs.djangoproject.com/en/dev/ref/files/file/



photo.image.save(name, File(open(content[0])), save=True)

photo.image.save(name, File(open(content[0])), save=True)

Combining what Chris Adams and Stan said and updating things to work on Python 3, if you install Requests you can do something like this:

from urllib.parse import urlparse
import requests
from django.core.files.base import ContentFile
from myapp.models import Photo

img_url = 'http://www.example.com/image.jpg'
name = urlparse(img_url).path.split('/')[-1]

photo = Photo() # set any other fields, but don't commit to DB (ie. don't save())

response = requests.get(img_url)
if response.status_code == 200:
    photo.image.save(name, ContentFile(response.content), save=True)

More relevant docs in Django's ContentFile documentation and Requests' file download example.

ImageField is just a string, a path relative to your MEDIA_ROOT setting. Just save the file (you might want to use PIL to check it is an image) and populate the field with its filename.

So it differs from your code in that you need to save the output of your urllib.urlopen to file (inside your media location), work out the path, save that to your model.

I do it this way on Python 3, which should work with simple adaptations on Python 2. This is based on my knowledge that the files I’m retrieving are small. If yours aren’t, I’d probably recommend writing the response out to a file instead of buffering in memory.

BytesIO is needed because Django calls seek() on the file object, and urlopen responses don’t support seeking. You could pass the bytes object returned by read() to Django's ContentFile instead.

from io import BytesIO
from urllib.request import urlopen

from django.core.files import File


# url, filename, model_instance assumed to be provided
response = urlopen(url)
io = BytesIO(response.read())
model_instance.image_field.save(filename, File(io))

Recently I use the following approach within python 3 and Django 3, maybe this might be interesting for others aswell. It is similar to Chris Adams solution but for me it did not work anymore.

import urllib.request
from django.core.files.uploadedfile import SimpleUploadedFile
from urllib.parse import urlparse

from demoapp import models


img_url = 'https://upload.wikimedia.org/wikipedia/commons/f/f7/Stack_Overflow_logo.png'
basename = urlparse(img_url).path.split('/')[-1]
tmpfile, _ = urllib.request.urlretrieve(img_url)

new_image = models.ModelWithImageOrFileField()
new_image.title = 'Foo bar'
new_image.file = SimpleUploadedFile(basename, open(tmpfile, "rb").read())
new_image.save()

Just discovered that you don't have to generate a temporary file:

Stream url content directly from django to minio

I have to store my files in minio and have django docker containers without much disk space and need to download big video files, so this was really helpful to me.

Its been almost 11 years since the question and the most reputed answer has been posted. Thanks To @chris-adams for the response. I am Just reposting the same answer along with the updated packages and support.

#! /usr/bin/python3
# lib/utils.py

import urllib3                                          # http Request Package.
from typing import Optional

from django.core.files import File                      # Handle Files in Django
from django.core.files.temp import NamedTemporaryFile   # handling temporary files.


def fetch_image(url: str, instance: models.Model, field: str, name: Optional[str]=None):
    """
    fetch_image Fetches an image URL and adds it to the model field.
    the parameter instance does not need to be a saved instance. 

    :url: str = A valid image URL.
    :instance: django.db.models.Model = Expecting a model with image field or file field. 
    :field: str = image / file field name as string; 
    [name:str] = Preferred file name, such as product slug or something.  

    :return: updated instance as django.db.models.Model, status of updation as bool.
    
    """
    
    conn = urllib3.PoolManager()
    response = conn.request('GET', url)
    if response.status <> 200:
        print("[X] 404! IMAGE NOT FOUND")
        print(f"TraceBack: {url}")
        return instance, False
    
    file_obj = NamedTemporaryFile(delete=True)
    file_obj.write( response.data )
    file_obj.flush()

    img_format = url.split('.')[-1]
    
    if name is None:
        name = url.split('/')[-1]
    
    if not name.endswith(img_format):
        name += f'.{img_format}'
    
    django_file_obj = File(file_obj)
    (getattr(instance, field)).save(name, django_file_obj)
    return instance, True
    

Tested with Django==2.2.12 in Python 3.7.5

if __name__ == '__main__':
    instance = ProductImage()
    url = "https://www.publicdomainpictures.net/pictures/320000/velka/background-image.png"
    instance, saved = fetch_image(url, instance, field='banner_image', name='intented-image-slug')
    status = ["FAILED! ", "SUCCESS! "][saved]
    print(status, instance.banner_image and instance.banner_image.path)
    instance.delete()

this is the right and working way

class Product(models.Model):
    upload_path = 'media/product'
    image = models.ImageField(upload_to=upload_path, null=True, blank=True)
    image_url = models.URLField(null=True, blank=True)

    def save(self, *args, **kwargs):
        if self.image_url:
            import urllib, os
            from urlparse import urlparse
            filename = urlparse(self.image_url).path.split('/')[-1]
            urllib.urlretrieve(self.image_url, os.path.join(file_save_dir, filename))
            self.image = os.path.join(upload_path, filename)
            self.image_url = ''
            super(Product, self).save()